Solve this following


Let $x=4$ be a directrix to an ellipse whose

centre is at the origin and its eccentricity is $\frac{1}{2}$.

If $P(1, \beta), \beta>0$ is a point on this ellipse, then the equation of the normal to it at $\mathrm{P}$ is :-



  1. $7 x-4 y=1$

  2. $4 x-2 y=1$

  3. $4 x-3 y=2$

  4. $8 x-2 y=5$

Correct Option: , 2


Ellipse : $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

$\operatorname{directrix}: x=\frac{a}{e}=4 \& e=\frac{1}{2}$

$\Rightarrow a=2 \& b^{2}=a^{2}\left(1-e^{2}\right)=3$

$\Rightarrow \quad$ Ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$

$P$ is $\left(1, \frac{3}{2}\right)$

Normal is : $\frac{4 x}{1}-\frac{3 y}{3 / 2}=4-3$

$\Rightarrow 4 x-2 y=1$


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